/*
Problem Description
Given two rectangles and the coordinates of two points on the diagonals of each rectangle,you have to calculate the area of the intersected part of two rectangles. its sides are parallel to OX and OY .
Input
Input The first line of input is 8 positive numbers which indicate the coordinates of four points that must be on each diagonal.The 8 numbers are x1,y1,x2,y2,x3,y3,x4,y4.That means the two points on the first rectangle are(x1,y1),(x2,y2);the other two points on the second rectangle are (x3,y3),(x4,y4).
Output
Output For each case output the area of their intersected part in a single line.accurate up to 2 decimal places.
Sample Input
1.00 1.00 3.00 3.00 2.00 2.00 4.00 4.00
5.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50
Sample Output
1.00
56.25
--------------------------------------------------------------
假设2个矩形的左下和右上 分别为 (x1,y1),(x2,y2) 和 (x3,y3),(x4,y4)
 HDU <wbr> <wbr>2056 <wbr> <wbr>(矩形重叠)Rectangles
则且仅当
(x4>x1&&y4>y1&&x3<x2&&y3<y2)
即: 矩形B的右上点>矩形A的左下点&&矩形B的左下点<矩形A的右上点
成立,2个矩形碰撞
现在分析不碰撞的情况
把求面积转化为求宽高
不难得到 宽(l)=MIN(x2,x4)-MAX(x1,x3);
             高(h)=MIN(y2,y4)-MAX(y1,y3);
于是乎，便可求得其解
最最该注意的是题中给出的对角线上的点是主对角线或副对角线上的
 */
package com.yuan.algorithms.training20150721;

import java.util.Scanner;

public class 根据坐标求俩矩形相交的面积 {

	static double x1, y1, x2, y2, x3, y3, x4, y4;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			x1 = sc.nextDouble();
			y1 = sc.nextDouble();
			x2 = sc.nextDouble();
			y2 = sc.nextDouble();
			x3 = sc.nextDouble();
			y3 = sc.nextDouble();
			x4 = sc.nextDouble();
			y4 = sc.nextDouble();
			change();
			double width = Math.min(x2, x4) - Math.max(x1, x3);
			double height = Math.min(y2, y4) - Math.max(y1, y3);
			if (width < 0 || height < 0) {
				System.out.println("0.00");
			} else {
				System.out.printf("%.2f", width * height);
				System.out.println();
			}
		}
	}

	/**
	 * 将对角线的坐标转换为主对角线的形式
	 */
	private static void change() {
		double t;
		if (x1 > x2) {
			t = x1;
			x1 = x2;
			x2 = t;
		}
		if (y1 > y2) {
			t = y1;
			y1 = y2;
			y2 = t;
		}
		if (x3 > x4) {
			t = x3;
			x3 = x4;
			x4 = t;
		}
		if (y3 > y4) {
			t = y3;
			y3 = y4;
			y4 = t;
		}
	}
	
}
